AMC12 2010 B

AMC12 2010 B · Q22

AMC12 2010 B · Q22. It mainly tests Circle theorems, Geometry misc.

Let $ABCD$ be a cyclic quadrilateral. The side lengths of $ABCD$ are distinct integers less than $15$ such that $BC\cdot CD=AB\cdot DA$. What is the largest possible value of $BD$?

设 $ABCD$ 是一个循环四边形。四边形 $ABCD$ 的边长是互不相同的小于 $15$ 的整数，且满足 $BC\cdot CD=AB\cdot DA$。$BD$ 的最大可能值是多少？

(A) \sqrt{\frac{325}{2}} \sqrt{\frac{325}{2}}

(B) \sqrt{185} \sqrt{185}

(D) \sqrt{\frac{425}{2}} \sqrt{\frac{425}{2}}

(E) \sqrt{\frac{533}{2}} \sqrt{\frac{533}{2}}

Answer

Correct choice: (D)

正确答案：（D）

Solution

Let $AB = a$, $BC = b$, $CD = c$, and $AD = d$. We see that by the Law of Cosines on $\triangle ABD$ and $\triangle CBD$, we have: $BD^2 = a^2 + d^2 - 2ad\cos{\angle BAD}$. $BD^2 = b^2 + c^2 - 2bc\cos{\angle BCD}$. We are given that $ad = bc$ and $ABCD$ is a cyclic quadrilateral. As a property of cyclic quadrilaterals, opposite angles are supplementary so $\angle BAD = 180 - \angle BCD$, therefore $\cos{\angle BAD} = -\cos{\angle BCD}$. So, $2ad\cos{\angle BAD} = -2bc\cos{\angle BCD}$. Adding, we get $2BD^2 = a^2 + b^2 + c^2 + d^2$. We now look at the equation $ad = bc$. Suppose that $a = 14$. Then, we must have either $b$ or $c$ equal $7$. Suppose that $b = 7$. We let $d = 6$ and $c = 12$. $2BD^2 = 196 + 49 + 36 + 144 = 425$, so our answer is $\boxed{\textbf{(D)} \sqrt{\frac{425}{2}}}$.

设 $AB = a$, $BC = b$, $CD = c$, 且 $AD = d$。由余弦定理分别作用于 $\triangle ABD$ 和 $\triangle CBD$，有： $BD^2 = a^2 + d^2 - 2ad\cos{\angle BAD}$。 $BD^2 = b^2 + c^2 - 2bc\cos{\angle BCD}$。已知 $ad = bc$ 且 $ABCD$ 为循环四边形。循环四边形的性质表明对角互补，所以 $\angle BAD = 180 - \angle BCD$，因此 $\cos{\angle BAD} = -\cos{\angle BCD}$。于是 $2ad\cos{\angle BAD} = -2bc\cos{\angle BCD}$。两式相加得 $2BD^2 = a^2 + b^2 + c^2 + d^2$。现在考虑方程 $ad = bc$。设 $a = 14$。则 $b$ 或 $c$ 必有一个等于 $7$。设 $b = 7$。取 $d = 6$ 且 $c = 12$。 $2BD^2 = 196 + 49 + 36 + 144 = 425$，所以答案是 $\boxed{\textbf{(D)} \sqrt{\frac{425}{2}}}$。

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